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The given equation is:

\[
\frac{3x}{x-2} + \frac{5}{x+2} = 1
\]

### Step 1: Eliminate the denominators
To solve this equation, we'll first eliminate the fractions by multiplying through by the least common denominator (LCD). The LCD for \(x-2\) and \(x+2\) is \((x-2)(x+2)\).

Multiply the entire equation by \((x-2)(x+2)\):

\[
(x-2)(x+2) \left( \frac{3x}{x-2} + \frac{5}{x+2} \right) = (x-2)(x+2) \cdot 1
\]

Simplifying both terms:

\[
3x(x+2) + 5(x-2) = (x-2)(x+2)
\]

### Step 2: Expand the terms
Now expand each side of the equation:

\[
3x(x+2) = 3x^2 + 6x
\]
\[
5(x-2) = 5x - 10
\]
\[
(x-2)(x+2) = x^2 - 4 \quad \text{(difference of squares)}
\]

Substituting these into the equation gives:

\[
3x^2 + 6x + 5x - 10 = x^2 - 4
\]

### Step 3: Simplify the equation
Combine like terms on the left side:

\[
3x^2 + 11x - 10 = x^2 - 4
\]

### Step 4: Move all terms to one side
Subtract \(x^2 - 4\) from both sides to set the equation to zero:

\[
3x^2 + 11x - 10 - x^2 + 4 = 0
\]

Simplifying:

\[
2x^2 + 11x - 6 = 0
\]

### Step 5: Solve the quadratic equation
Now we solve the quadratic equation \(2x^2 + 11x - 6 = 0\). Using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \(a = 2\), \(b = 11\), and \(c = -6\).

First, calculate the discriminant:

\[
\Delta = b^2 - 4ac = 11^2 - 4(2)(-6) = 121 + 48 = 169
\]

Now solve for \(x\):

\[
x = \frac{-11 \pm \sqrt{169}}{2(2)} = \frac{-11 \pm 13}{4}
\]

So the two solutions are:

\[
x_1 = \frac{-11 + 13}{4} = \frac{2}{4} = \frac{1}{2}
\]
\[
x_2 = \frac{-11 - 13}{4} = \frac{-24}{4} = -6
\]

### Step 6: Check for extraneous solutions
We must check that these solutions do not make the denominators zero. The denominators are \(x-2\) and \(x+2\), so the values \(x = 2\) and \(x = -2\) are not allowed. Neither \(x = \frac{1}{2}\) nor \(x = -6\) makes the denominators zero, so both solutions are valid.

### Final Answer:
The solutions are:
\[
x = \frac{1}{2}, \quad x = -6
\]

Would you like further clarification or details?

Here are 5 follow-up questions based on this problem:
1. How is the least common denominator determined for fractions?
2. What is the difference of squares and how is it applied here?
3. Can the quadratic equation be solved without using the quadratic formula?
4. Why is it important to check for extraneous solutions in rational equations?
5. How can you graphically interpret the solutions of a quadratic equation?

**Tip:** Always verify your solutions by substituting them back into the original equation to ensure they do not cause division by zero or other inconsistencies.

Solve the equation: (3x / (x - 2)) + (5 / (x + 2)) = 1

Learn how to solve the rational equation (3x/(x-2)) + (5/(x+2)) = 1 step-by-step using algebraic techniques such as finding the least common denominator and solving a resulting quadratic equation. This problem involves the quadratic formula and is suitable for students in Grades 9-11.

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